A maybe trivial question about fiber bundles (I'm not an expert, and I didn't find quickly an answer looking here and there). Suppose you are given fiber bundle $p\colon E\to M$, where
$E,M$ are smooth manifolds and $p$ is smoothly locally trivial. Also suppose that the fiber $F$ of the bundle is smoothly contractible. Is that true that $p$ admits a smooth section?

3 Answers
3

In fact you only need that the fiber is contractible not smoothly contractible. Take any continuous section $s_0 \colon B \to E$. cover $B$ by open sets $U_i$ such that the bundle is trivializable over each $U_i$, also make sure that the closure of each $U_i$ is compact and that the cover is locally finite.

Furthermore, give $E$ any complete Riemannian structure. This provides each fiber $E_x$ with an induced Riemannian structure, which is also complete. We use this to define the obvious supremum distance between any two sections of $E$ over any sub-space of $B$. We may also construct a continuous map $r \colon E \to \mathbb{R}_+$ such that the ball of radius $r(x)$ and center $x$ in each fiber is geodesically convex.

The construction now goes in 2 steps:

1) local construction: for each $i$ we may find a smooth section $s \colon U_i \to E$ such that the supremum distance defined above is smaller than $r$ on all the points of $S_0$ restricted to $U_i$. This is easy and follows from smooth approximation of any function from $U_i \to F$ defining a section $U_i \to U_i \times F$.

2) global construction: use a partion of unity to get a global construction. This is now possible because we were carefull enough to create the smooth local sections such that they lie in a geodesically convex neighborhood.

While finishing this Andrew Stacey posted a similar answer, but it seemed a waste not to poste this also. Especially since we are focussing on different details.

Nice answer. I always feel that appealing to a Riemannian structure is overkill for results that are differential topological in nature, but I know that I'm in the minority on that one! And also adding a Riemannian structure can provide a fast shortcut for many proofs so it's useful to know what one can do with it.
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Loop SpaceMay 25 '10 at 14:35

I agree with Andrew Stacey that the most "natural" setting where looking for a proof is differential topology. On the other hand, I still remember with pain when a professor of mine proved that every paracompact smooth manifold admits a good cover (i.e. a locally finite cover made of contractible open sets whose arbitrary intersections are all contractible) without using any auxiliary Riemannian metric! Since then, I developed some sympathy for Riemannian methods in differential topology (and I am actually a fan of injectivity radius and related issues).
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Roberto FrigerioMay 25 '10 at 15:40

I tend to work with infinite dimensional stuff where Riemannian structures aren't always available, and yet the differential topology still works, so I like methods that don't use the metric as they generalise more easily.
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Loop SpaceMay 25 '10 at 20:10

Yes. This follows from smooth paracompactness of $M$. Basically, you choose local trivialisations over a suitable cover and then on the intersections you smoothly interpolate between the different choices. By choosing your cover carefully, at each stage you are interpolating between a finite number of choices so the interpolation amounts to smoothly filling in a simplex, which by assumption you can do. (In slightly more detail, but only slightly, once you've filled in the simplex, you use a partition of unity to specify where in the simplex you should be. Pictorially, if you imagine a line at $y = 0$ and one at $y = 1$, then between $x=0$ and $x=1$ you must get from one line to the other. "Filling in the simplex" corresponds to filling in the rectangle $(0,0) -- (1,0) -- (1,1) -- (0,1)$. A partition of unity in this case is just a smooth function $[0,1] \to [0,1]$ which is $0$ near $0$ and $1$ near $1$. The graph of this defines the interpolation between the two lines and thus the way to get from the line $y = 0$ to the one at $y= 1$.)

It is interesting to note that, if the base manifold M is not paracompact, there can be situations in which no sections exist. A standard example is M = the long line, and P = the positive half of its tangent bundle.
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André HenriquesMay 25 '10 at 16:55